Question #173478

7(b) Let X X Xn

, , , 1 2 K be random sample of size n from a distribution with probability 

density function 

θ < < θ >

=

θ−

,0 else where.

0, ,1 0

( )0,;

1 X X

f X Obtain a maximum likeyhood 

estimator of θ.


1
Expert's answer
2021-03-30T06:27:27-0400

Given probability distribution function is-


f(x;θ)=θ21e(xθ1)θ2f(x;θ)=θ^{−1}_2e^{−\dfrac{(x−θ_1)}{θ_2}}


. The likelihood function is the density function regarded as a function of θ.


L(θx)=f(xθ),θΘ.(1)L(θ|x) = f(x|θ), θ ∈ Θ. (1)


The maximum likelihood estimator (MLE),


θ^(x)=argmaxL(θx).\hat{θ}(x) = arg maxL(θ|x).


for x>θ1,θ2>0x>θ_1,θ_2>0 Likelihood:


L(θ)=L(θX1,X2,,Xn)=i=1ne(xiθ1)θ2θ2L(θ)=L(θ∣X_1,X_2,…,X_n)=∏_{i=1}^n\dfrac{e^{\dfrac{−(x_i−θ_1)}{θ_2}}}{θ_2}

factored out θ2θ_2 as well as turned to product into a summation to get:


=1θ2nei=1n(xiθ1)θ2=\dfrac{1}{θ^n_2}e^{∑^n_{i=1}}\dfrac{−(x_i−θ_1)}{θ_2}


(θ)=ln(L(θ))=1θ2nei=1n(xiθ1)θ2ℓ(θ)=ln(L(θ))=\dfrac{1}{θ^n_2}e^{∑^n_{i=1}}\dfrac{−(x_i−θ_1)}{θ_2}


=ln(θ2n)+i=1n(xiθ1)θ2=ln(θ^{−n}_2)+∑_{i=1}^n−\dfrac{(x_i−θ_1)}{θ_2}


   =nln(θ2)i=1nxinθ1θ2=−nln(θ_2)−\dfrac{∑^n_{i=1}x_i−nθ_1}{θ_2}


    =nln(θ2)i=1nxiθ2+nθ1θ2=−nln(θ_2)−\dfrac{∑^n_{i=1}x_i}{θ_2}+\dfrac{nθ_1}{\theta_2}


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS